Question

A particle is moving along a line parallel to x-axis at a distance d from x-axis, with constant speed u. Find its radial acceleration using polar coordinates

Answer 1

Suppose the particle has coordinates $(R, \phi)$ at some moment $t$. The formula for radial acceleration is:

$$a = \frac{v^2}{R}$$

where $v$ is a tangential velocity.

As $v$ is perpendicular to OB (because it is tangential velocity), we get from the scheme that:

$$v = u \sin \phi$$

Then

$$a(R, \phi) = \frac{u^2 \sin^2 \phi}{R}$$

If we assume that at the moment $t_0 = 0$ the particle was at the point A (just in opposite to the origin), then it was at the point B at some moment $t$ and $AB = ut$.

Then we get:

$$R = \sqrt{u^2 t^2 + d^2}$$

and

$$\tan \phi = \frac {d}{u t}$$

So, $\sin^2\phi = 1 - \cos^2\phi = 1 - \frac{1}{1 + \tan^2\phi} = \frac{\tan^2\phi}{1 + \tan^2\phi} = \frac{d^2}{u^2t^2 + d^2}.$

Finally,

$$a (t) = \frac {u ^ {2}}{\sqrt {u ^ {2} t ^ {2} + d ^ {2}}} * \frac {d ^ {2}}{u ^ {2} t ^ {2} + d ^ {2}} = \frac {u ^ {2} d ^ {2}}{(u ^ {2} t ^ {2} + d ^ {2}) ^ {\frac {3}{2}}}$$